LeetCode——Regular Expression Matching

来源：互联网发布：小清新调色思路知乎编辑：程序博客网时间：2024/05/20 23:40

Implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.'*' Matches zero or more of the preceding element.The matching should cover the entire input string (not partial).The function prototype should be:bool isMatch(const char *s, const char *p)Some examples:isMatch("aa","a") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

实现支持 ' . '和 ' * '的正则表达式。

' . ' 匹配任何单字符。

' * '匹配0或多个前向元素。

使用递归进行判断。总体上可以分成两种情况，一种是以 ' * ‘开头的，另一种不是。

public class RegularExpressionMatching {public static void main(String[] args) {System.out.println(isMatch("aa","a"));System.out.println(isMatch("aa","aa"));System.out.println(isMatch("aaa","aa"));System.out.println(isMatch("aa", "a*"));System.out.println(isMatch("aa", ".*"));System.out.println(isMatch("ab", ".*"));System.out.println(isMatch("aab", "c*a*b"));}public static boolean isMatch(String s,String p){if(p.length() == 0)return s.length() == 0;if(p.length() == 1 || p.charAt(1) != '*'){if(s.length()  < 1 || (p.charAt(0) != '.' && s.charAt(0) != p.charAt(0)))return false;return isMatch(s.substring(1),p.substring(1));}else{int i = -1;while(i < s.length() && (i < 0 || p.charAt(0) == '.' || p.charAt(0) == s.charAt(i))){if(isMatch(s.substring(i+1),p.substring(2)))return true;i++;}return false;}}}

Reference：http://www.programcreek.com/2012/12/leetcode-regular-expression-matching-in-java/

0 0